【前缀和】Day 8 提高组模拟C组 T1 七夕祭

题目大意

一张图上有若干个点，每个点都可以与相邻格子交换，问最少要转换几次使得这些点都在同一行或同一列

解题思路

首先这是一个和均分纸牌几乎一毛一样的问题，我们可以通过排序然后走一遍环形均分纸牌，时间复杂度:$O(n^2)$，得分70
然后可以通过前缀和优化至$O(nlogn)$，因为我们需要合并的次数其实就是每个前缀和与其中位数的差，求和即可

代码

#include<cstdio>
#include<algorithm>
#define LL long long
using namespace std;int h[100001],l[100001],T,n,m,x,y;LL ansh,ansl,sh[100001],sl[100001];
LL f;
char c;
inline LL read()//读入优化
{
    f=0;
    while(c=getchar(),c<=47||c>=58);f=(f<<3)+(f<<1)+c-48;
    while(c=getchar(),c>=48&&c<=57) f=(f<<3)+(f<<1)+c-48;
    return f;
}
void write(LL x){if(x>9) write(x/10);putchar(x%10+48);return;}//输出优化
int main()
{
    n=read();m=read();T=read();
    if(T%n&&T%m) return printf("impossible")&0;//不能移动
    for(int i=1;i<=T;i++)
    {
        x=read();y=read();
        h[x]++;l[y]++;
    }
    for(int i=1;i<=n;i++) sh[i]=sh[i-1]+h[i]-T/n;
    for(int i=1;i<=m;i++) sl[i]=sl[i-1]+l[i]-T/m;
    sort(sh+1,sh+1+n);sort(sl+1,sl+1+m);
    for(int i=1;i<=n;i++) ansh+=abs(sh[(n+1)>>1]-sh[i]);
    for(int i=1;i<=m;i++) ansl+=abs(sl[(m+1)>>1]-sl[i]);//求值
    if(!(T%n)&&!(T%m)) //行列皆可
    {
        printf("both ");
        write(ansh+ansl);//输出
    }
    else
    if(!(T%n)) printf("row "),write(ansh);//只有行
    else
    if(!(T%m)) printf("column "),write(ansl);//只有列
}

正解

#include<cstdio>
using namespace std;int k;
int main()
{
    scanf("%d",&k);
    if(k==2) printf("4 0011");
    if(k==3) printf("8 00010111");
    if(k==4) printf("16 0000100110101111");
    if(k==5) printf("32 00000100011001010011101011011111");
    if(k==6) printf("64 0000001000011000101000111001001011001101001111010101110110111111");
    if(k==7) printf("128 00000001000001100001010000111000100100010110001101000111100100110010101001011100110110011101001111101010110101111011011101111111");
    if(k==8) printf("256 0000000010000001100000101000001110000100100001011000011010000111100010001001100010101000101110001100100011011000111010001111100100101001001110010101100101101001011110011001101010011011100111011001111010011111101010101110101101101011111011011110111011111111");
    if(k==9) printf("512 00000000010000000110000001010000001110000010010000010110000011010000011110000100010000100110000101010000101110000110010000110110000111010000111110001000110001001010001001110001010010001010110001011010001011110001100110001101010001101110001110010001110110001111010001111110010010010110010011010010011110010100110010101010010101110010110110010111010010111110011001110011010110011011010011011110011101010011101110011110110011111010011111110101010110101011110101101110101110110101111110110110111110111011110111111111");
    if(k==10) printf("1024 0000000000100000000110000000101000000011100000010010000001011000000110100000011110000010001000001001100000101010000010111000001100100000110110000011101000001111100001000010001100001001010000100111000010100100001010110000101101000010111100001100010000110011000011010100001101110000111001000011101100001111010000111111000100010100010001110001001001000100101100010011010001001111000101001100010101010001010111000101100100010110110001011101000101111100011000110010100011001110001101001000110101100011011010001101111000111001100011101010001110111000111100100011110110001111101000111111100100100110010010101001001011100100110110010011101001001111100101001010011100101010110010101101001010111100101100110010110101001011011100101110110010111101001011111100110011010011001111001101010100110101110011011011001101110100110111110011100111010110011101101001110111100111101010011110111001111101100111111010011111111010101010111010101101101010111110101101011011110101110111010111101101011111110110110111011011111101110111110111101111111111");
    if(k==11) printf("2048 00000000000100000000011000000001010000000011100000001001000000010110000000110100000001111000000100010000001001100000010101000000101110000001100100000011011000000111010000001111100000100001000001000110000010010100000100111000001010010000010101100000101101000001011110000011000100000110011000001101010000011011100000111001000001110110000011110100000111111000010000110000100010100001000111000010010010000100101100001001101000010011110000101000100001010011000010101010000101011100001011001000010110110000101110100001011111000011000110000110010100001100111000011010010000110101100001101101000011011110000111000100001110011000011101010000111011100001111001000011110110000111110100001111111000100010010001000101100010001101000100011110001001001100010010101000100101110001001100100010011011000100111010001001111100010100011000101001010001010011100010101001000101010110001010110100010101111000101100110001011010100010110111000101110010001011101100010111101000101111110001100011100011001001000110010110001100110100011001111000110100110001101010100011010111000110110010001101101100011011101000110111110001110010100011100111000111010010001110101100011101101000111011110001111001100011110101000111101110001111100100011111011000111111010001111111100100100101001001001110010010101100100101101001001011110010011001100100110101001001101110010011101100100111101001001111110010100101100101001101001010011110010101001100101010101001010101110010101101100101011101001010111110010110011100101101011001011011010010110111100101110011001011101010010111011100101111011001011111010010111111100110011011001100111010011001111100110100111001101010110011010110100110101111001101101010011011011100110111011001101111010011011111100111001111001110101010011101011100111011011001110111010011101111100111101011001111011010011110111100111110101001111101110011111101100111111101001111111110101010101101010101111010101101110101011101101010111111010110101110101101101101011011111010111011110101111011101011111011010111111110110110111101101110111011011111110111011111101111011111011111111111");
}